Solution:

The given eq. are:
$\begin{array}{l}
x+\frac{6}{y}=6 \quad \ldots \ldots \text {(i) } \\
3 x-\frac{8}{y}=5 \ldots \ldots \text { (ii) }
\end{array}$
Putting $\frac{1}{y}=\mathrm{v}$, we obtain:
$\begin{array}{l}
x+6 v=6 \ldots \ldots \text { (iii) } \\
3 x-8 v=5 \ldots \ldots \text { (iv) }
\end{array}$
Upon multiplying eq.(iii) by 4 and eq.(iv) by 3 , we obtain:
$\begin{array}{l}
4 x+24 v=24 \quad \ldots \ldots \ldots(v) \\
9 x-24 v=15 \quad \ldots \ldots \text { (vi) }
\end{array}$
On adding eq.(v) from eq.(vi), we obtain:
$13 x=39 \Rightarrow x=3$
On substituting $\mathrm{x}=3$ in equation(i), we obtain:
$\begin{array}{l}
3+\frac{6}{y}=6 \\
\Rightarrow \frac{6}{y}=(6-3)=3 \Rightarrow 3 y=6 \Rightarrow y=2
\end{array}$
As a result, the required solution is $\mathrm{x}=3$ and $\mathrm{y}=2$.